Уравнение вида a sin x + b cos x = c. Задача 16

ОТВЕТ: \(\arcsin \dfrac{3}{{2\sqrt 5 }} + \arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k;\,\,\,\pi -\arcsin \dfrac{3}{{2\sqrt 5 }} + \arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k;\,\,\,k \in Z.\) 

\(2\sin x-4\cos x = 3.\)

Уравнение вида:  \(a\sin x + b\cos x = c,\;\;\)где \(\;a = 2,\,\,\,\,b = -4,\;\;c = 3.\)

Разделим обе части исходного уравнения на  \(\sqrt {{a^2} + {b^2}} ,\;\) то есть на  \(\sqrt {{2^2} + {{\left( {-4} \right)}^2}}  = 2\sqrt 5 .\)

\(\dfrac{1}{{\sqrt 5 }}\sin x-\dfrac{2}{{\sqrt 5 }}\cos x = \dfrac{3}{{2\sqrt 5 }}.\)

Так как  \({\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2} + {\left( {\dfrac{2}{{\sqrt 5 }}} \right)^2} = 1,\)  то пусть  \(\cos \varphi  = \dfrac{1}{{\sqrt 5 }},\,\,\,\sin \varphi  = \dfrac{2}{{\sqrt 5 }}.\)  Тогда уравнение примет вид:

\(\dfrac{1}{{\sqrt 5 }}\sin x-\dfrac{2}{{\sqrt 5 }}\cos x = \dfrac{3}{{2\sqrt 5 }}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\cos \varphi \sin x-\sin \varphi \cos x = \dfrac{3}{{2\sqrt 5 }}\,\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\sin \left( {x-\varphi } \right) = \dfrac{3}{{2\sqrt 5 }}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-\varphi  = \arcsin \dfrac{3}{{2\sqrt 5 }} + 2\pi k,\,\,\,\,\,\,}\\{x-\varphi  = \pi -\arcsin \dfrac{3}{{2\sqrt 5 }} + 2\pi k}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = \arcsin \dfrac{3}{{2\sqrt 5 }} + \varphi  + 2\pi k,\,\,\,\,\,\,\,\,\,}\\{x = \pi -\arcsin \dfrac{3}{{2\sqrt 5 }} + \varphi  + 2\pi k,}\end{array}} \right.\,\,\,\,k \in Z.\)

Так как  \(\sin \varphi  = \dfrac{2}{{\sqrt 5 }},\)  то   \(\varphi  = \arcsin \dfrac{2}{{\sqrt 5 }}\)  и  

\(\left[ {\begin{array}{*{20}{c}}{x = \arcsin \dfrac{3}{{2\sqrt 5 }} + \arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k,\,\,\,\,\,\,\,\,\,}\\{x = \pi -\arcsin \dfrac{3}{{2\sqrt 5 }} + \arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k,}\end{array}} \right.\,\,\,\,\,k \in Z.\)

Ответ:  \(\arcsin \dfrac{3}{{2\sqrt 5 }} + \arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k;\,\,\,\pi -\arcsin \dfrac{3}{{2\sqrt 5 }} + \arcsin \dfrac{2}{{\sqrt 5 }} + 2\pi k;\,\,\,k \in Z.\)